To prove that a positive integer n is even if and only if 7n + 4 is even, we will break this down into two parts: the ‘if’ part and the ‘only if’ part.
If n is even, then 7n + 4 is even
Suppose n is an even positive integer. By definition, this means there exists some integer k such that n = 2k.
Now, substituting this into the expression 7n + 4:
7n + 4 = 7(2k) + 4 = 14k + 4 = 2(7k + 2)
Since the expression 7k + 2 is an integer (as both 7k and 2 are integers), we can see that 7n + 4 is clearly a multiple of 2. Hence, 7n + 4 is even.
If 7n + 4 is even, then n is even
Now, let’s assume that 7n + 4 is an even integer. This means that there exists some integer m such that:
7n + 4 = 2m
Rearranging this gives:
7n = 2m – 4
Factoring out a 2 from the right side:
7n = 2(m – 2)
To isolate n, we divide both sides by 7:
n = (2(m – 2))/7
For n to be an integer, 2(m – 2) must be divisible by 7. Now we need to examine the form of 2(m – 2). Because it is an even number (since it is clearly a multiple of 2), if it is divisible by 7, we consider what kind of integer m – 2 could be.
Note that for n to be an integer, (m – 2) must be such that multiplying by 2 and dividing by 7 yields an integer. Thus, n is an integer whenever m – 2 is even, which implies that m must also be even. Therefore, resulting from this condition keeps m to be even, 7n + 4 being even indicates that n must also be even.
Thus, we have shown both directions of the proof: if n is even, then 7n + 4 is even; and if 7n + 4 is even, then n is even. Therefore, we conclude:
n is even if and only if 7n + 4 is even.